Optimal. Leaf size=229 \[ -\frac {\sqrt {3} \text {ArcTan}\left (\frac {\sqrt {3} \sqrt [3]{1+\left (-1-k^2\right ) x^2+k^2 x^4}}{2 \sqrt [3]{d}-2 \sqrt [3]{d} x+\sqrt [3]{1+\left (-1-k^2\right ) x^2+k^2 x^4}}\right )}{\sqrt [3]{d}}-\frac {\log \left (-\sqrt [3]{d}+\sqrt [3]{d} x+\sqrt [3]{1+\left (-1-k^2\right ) x^2+k^2 x^4}\right )}{\sqrt [3]{d}}+\frac {\log \left (d^{2/3}-2 d^{2/3} x+d^{2/3} x^2+\left (\sqrt [3]{d}-\sqrt [3]{d} x\right ) \sqrt [3]{1+\left (-1-k^2\right ) x^2+k^2 x^4}+\left (1+\left (-1-k^2\right ) x^2+k^2 x^4\right )^{2/3}\right )}{2 \sqrt [3]{d}} \]
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Rubi [F]
time = 1.55, antiderivative size = 0, normalized size of antiderivative = 0.00, number of steps
used = 0, number of rules used = 0, integrand size = 0, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.000, Rules used = {}
\begin {gather*} \int \frac {3+\left (1-2 k^2\right ) x-3 k^2 x^2+k^2 x^3}{\sqrt [3]{\left (1-x^2\right ) \left (1-k^2 x^2\right )} \left (-1+d-(1+2 d) x+\left (d+k^2\right ) x^2+k^2 x^3\right )} \, dx \end {gather*}
Verification is not applicable to the result.
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Rubi steps
\begin {align*} \int \frac {3+\left (1-2 k^2\right ) x-3 k^2 x^2+k^2 x^3}{\sqrt [3]{\left (1-x^2\right ) \left (1-k^2 x^2\right )} \left (-1+d-(1+2 d) x+\left (d+k^2\right ) x^2+k^2 x^3\right )} \, dx &=\frac {\left (\sqrt [3]{1-x^2} \sqrt [3]{1-k^2 x^2}\right ) \int \frac {3+\left (1-2 k^2\right ) x-3 k^2 x^2+k^2 x^3}{\sqrt [3]{1-x^2} \sqrt [3]{1-k^2 x^2} \left (-1+d-(1+2 d) x+\left (d+k^2\right ) x^2+k^2 x^3\right )} \, dx}{\sqrt [3]{\left (1-x^2\right ) \left (1-k^2 x^2\right )}}\\ &=\frac {\left (\sqrt [3]{1-x^2} \sqrt [3]{1-k^2 x^2}\right ) \int \left (\frac {1}{\sqrt [3]{1-x^2} \sqrt [3]{1-k^2 x^2}}+\frac {4-d+2 \left (1+d-k^2\right ) x-\left (d+4 k^2\right ) x^2}{\sqrt [3]{1-x^2} \sqrt [3]{1-k^2 x^2} \left (-1+d-(1+2 d) x+\left (d+k^2\right ) x^2+k^2 x^3\right )}\right ) \, dx}{\sqrt [3]{\left (1-x^2\right ) \left (1-k^2 x^2\right )}}\\ &=\frac {\left (\sqrt [3]{1-x^2} \sqrt [3]{1-k^2 x^2}\right ) \int \frac {1}{\sqrt [3]{1-x^2} \sqrt [3]{1-k^2 x^2}} \, dx}{\sqrt [3]{\left (1-x^2\right ) \left (1-k^2 x^2\right )}}+\frac {\left (\sqrt [3]{1-x^2} \sqrt [3]{1-k^2 x^2}\right ) \int \frac {4-d+2 \left (1+d-k^2\right ) x-\left (d+4 k^2\right ) x^2}{\sqrt [3]{1-x^2} \sqrt [3]{1-k^2 x^2} \left (-1+d-(1+2 d) x+\left (d+k^2\right ) x^2+k^2 x^3\right )} \, dx}{\sqrt [3]{\left (1-x^2\right ) \left (1-k^2 x^2\right )}}\\ &=\frac {x \sqrt [3]{1-x^2} \sqrt [3]{1-k^2 x^2} F_1\left (\frac {1}{2};\frac {1}{3},\frac {1}{3};\frac {3}{2};x^2,k^2 x^2\right )}{\sqrt [3]{\left (1-x^2\right ) \left (1-k^2 x^2\right )}}+\frac {\left (\sqrt [3]{1-x^2} \sqrt [3]{1-k^2 x^2}\right ) \int \left (\frac {\left (1-\frac {4}{d}\right ) d}{\sqrt [3]{1-x^2} \sqrt [3]{1-k^2 x^2} \left (1-d+(1+2 d) x-\left (d+k^2\right ) x^2-k^2 x^3\right )}+\frac {2 \left (-1-d+k^2\right ) x}{\sqrt [3]{1-x^2} \sqrt [3]{1-k^2 x^2} \left (1-d+(1+2 d) x-\left (d+k^2\right ) x^2-k^2 x^3\right )}+\frac {\left (d+4 k^2\right ) x^2}{\sqrt [3]{1-x^2} \sqrt [3]{1-k^2 x^2} \left (1-d+(1+2 d) x-\left (d+k^2\right ) x^2-k^2 x^3\right )}\right ) \, dx}{\sqrt [3]{\left (1-x^2\right ) \left (1-k^2 x^2\right )}}\\ &=\frac {x \sqrt [3]{1-x^2} \sqrt [3]{1-k^2 x^2} F_1\left (\frac {1}{2};\frac {1}{3},\frac {1}{3};\frac {3}{2};x^2,k^2 x^2\right )}{\sqrt [3]{\left (1-x^2\right ) \left (1-k^2 x^2\right )}}+\frac {\left ((-4+d) \sqrt [3]{1-x^2} \sqrt [3]{1-k^2 x^2}\right ) \int \frac {1}{\sqrt [3]{1-x^2} \sqrt [3]{1-k^2 x^2} \left (1-d+(1+2 d) x-\left (d+k^2\right ) x^2-k^2 x^3\right )} \, dx}{\sqrt [3]{\left (1-x^2\right ) \left (1-k^2 x^2\right )}}-\frac {\left (2 \left (1+d-k^2\right ) \sqrt [3]{1-x^2} \sqrt [3]{1-k^2 x^2}\right ) \int \frac {x}{\sqrt [3]{1-x^2} \sqrt [3]{1-k^2 x^2} \left (1-d+(1+2 d) x-\left (d+k^2\right ) x^2-k^2 x^3\right )} \, dx}{\sqrt [3]{\left (1-x^2\right ) \left (1-k^2 x^2\right )}}+\frac {\left (\left (d+4 k^2\right ) \sqrt [3]{1-x^2} \sqrt [3]{1-k^2 x^2}\right ) \int \frac {x^2}{\sqrt [3]{1-x^2} \sqrt [3]{1-k^2 x^2} \left (1-d+(1+2 d) x-\left (d+k^2\right ) x^2-k^2 x^3\right )} \, dx}{\sqrt [3]{\left (1-x^2\right ) \left (1-k^2 x^2\right )}}\\ \end {align*}
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Mathematica [F]
time = 41.41, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {3+\left (1-2 k^2\right ) x-3 k^2 x^2+k^2 x^3}{\sqrt [3]{\left (1-x^2\right ) \left (1-k^2 x^2\right )} \left (-1+d-(1+2 d) x+\left (d+k^2\right ) x^2+k^2 x^3\right )} \, dx \end {gather*}
Verification is not applicable to the result.
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Maple [F]
time = 0.02, size = 0, normalized size = 0.00 \[\int \frac {3+\left (-2 k^{2}+1\right ) x -3 k^{2} x^{2}+k^{2} x^{3}}{\left (\left (-x^{2}+1\right ) \left (-k^{2} x^{2}+1\right )\right )^{\frac {1}{3}} \left (-1+d -\left (1+2 d \right ) x +\left (k^{2}+d \right ) x^{2}+k^{2} x^{3}\right )}\, dx\]
Verification of antiderivative is not currently implemented for this CAS.
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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [F(-1)] Timed out
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Sympy [F(-1)] Timed out
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Mupad [F]
time = 0.00, size = -1, normalized size = -0.00 \begin {gather*} -\int \frac {3\,k^2\,x^2-k^2\,x^3+x\,\left (2\,k^2-1\right )-3}{{\left (\left (x^2-1\right )\,\left (k^2\,x^2-1\right )\right )}^{1/3}\,\left (d+k^2\,x^3+x^2\,\left (k^2+d\right )-x\,\left (2\,d+1\right )-1\right )} \,d x \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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